1. **Problem Statement:** We have three subsets of whole numbers from 1 to 20: - $F = \{2, 4, 6, 8, 10, 12, 14, 18\}$ - $G = \{1, 2, 3, 4, 5, 7, 9, 11, 16, 18\}$ - $H = \{1, 2, 3, 5, 7, 9, 14\}$ We need to find: (i) $F \cap G$ (ii) $(F \cup G)^c$ (iii) $(F \cap H)^c$ 2. **Formulas and Rules:** - Intersection $A \cap B$ is the set of elements common to both $A$ and $B$. - Union $A \cup B$ is the set of elements in $A$, or $B$, or both. - Complement $A^c$ is the set of elements not in $A$, relative to the universal set (here, whole numbers from 1 to 20). 3. **Step-by-step Solutions:** (i) Find $F \cap G$ (elements common to both $F$ and $G$): $F = \{2, 4, 6, 8, 10, 12, 14, 18\}$ $G = \{1, 2, 3, 4, 5, 7, 9, 11, 16, 18\}$ Common elements: $2, 4, 18$ So, $$F \cap G = \{2, 4, 18\}$$ (ii) Find $(F \cup G)^c$ (elements not in $F$ or $G$): First find $F \cup G$: $F \cup G = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 16, 18\}$ Universal set $U = \{1, 2, 3, ..., 20\}$ Elements not in $F \cup G$ are those in $U$ but not in $F \cup G$: $\{13, 15, 17, 19, 20\}$ So, $$(F \cup G)^c = \{13, 15, 17, 19, 20\}$$ (iii) Find $(F \cap H)^c$ (elements not in both $F$ and $H$): First find $F \cap H$: $F = \{2, 4, 6, 8, 10, 12, 14, 18\}$ $H = \{1, 2, 3, 5, 7, 9, 14\}$ Common elements: $2, 14$ So, $$F \cap H = \{2, 14\}$$ Now find complement relative to $U$: $$(F \cap H)^c = U - \{2, 14\} = \{1, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 17, 18, 19, 20\}$$ **Final answers:** (i) $F \cap G = \{2, 4, 18\}$ (ii) $(F \cup G)^c = \{13, 15, 17, 19, 20\}$ (iii) $(F \cap H)^c = \{1, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 17, 18, 19, 20\}$ These results can be visualized on a Venn diagram with three overlapping circles representing $F$, $G$, and $H$ within the universal set of numbers 1 to 20.